Constant Acceleration of Two Cars

Click For Summary

Homework Help Overview

The problem involves two cars traveling along a straight road, with one car maintaining a constant speed while the other is faster and initially behind. The discussion centers around determining how far the slower car will travel before being overtaken and the distance the faster car will be ahead after overtaking.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the concept of constant acceleration versus constant speed, questioning the relevance of acceleration in this context.
  • Some suggest formulating equations for the positions of the cars as functions of time, while others express confusion about the variables and units used in their calculations.
  • There are attempts to solve for the time it takes for the faster car to catch up, with varying degrees of clarity and correctness in the algebraic manipulations.
  • Questions arise regarding the interpretation of the variables and the setup of the equations, prompting requests for clarification and reasoning.

Discussion Status

The discussion is ongoing, with participants providing hints and guidance to each other. Some have made progress in understanding the relationships between the distances traveled by the cars, while others are still grappling with the algebra involved. There is no explicit consensus on the final approach or solution yet.

Contextual Notes

Participants note potential confusion regarding the definitions of variables and the implications of constant speed versus acceleration. There is also mention of the need to clarify units and the setup of the problem as it relates to the distances traveled by each car.

ff4930
Messages
46
Reaction score
0
[SOLVED] Constant Acceleration

Homework Statement



Two cars are traveling along a straight road. Car A maintains a constant speed of 80 km/h and car B maintains a constant speed of 110 km/hr. At t=0, car B is 45 km behind Car A.
A) How much farther will car A travel before Car b overtakes it?
B)How much ahead of A will B be in 30 seconds after it overtakes A?


Homework Equations





The Attempt at a Solution



I am stumped. I read and reread the section on my textbook about constant acceleration and the material is so confusing. Can someone please hint me or point me to a direction of understanding this please. Thanks.
 
Physics news on Phys.org
Why constant acceleration? Acceleration is 0 as you only have constant speed right?
 
Come up with expressions for the position of each car as a function of time. I'd call the location of Car b at t = 0 to be the origin. (So Car A is at 45 km at t = 0.)

There's no acceleration mentioned, so this should be easy.
 
110x = 80x+45
x=.5?
is this the right approach? I am really lost.
 
ff4930 said:
110x = 80x+45
x=.5?
is this the right approach? I am really lost.
Your equation is right, but how did you get x = .5? And what are the units? For that matter, what is x? What have you solved for here?

If you explain your reasoning a little more, we can help you better.
 
At the time instant where two cars meet up, the distance traveled by Car B will be 45 more than Car A...
v = x/t is the main equation for this type of question.
 
belliott4488 said:
Your equation is right, but how did you get x = .5? And what are the units? For that matter, what is x? What have you solved for here?

If you explain your reasoning a little more, we can help you better.
.5 hr? it will take .5 hr for Car B to catch up to Car A?

So Car A will travel another 40 km before Car B catches up.
 
Last edited:
Okay, so x is time (it's more conventional to use "t" for a time interval and "x" for distance, but it's not necessary). I think you just need to be a little more careful in solving your equation for x - you just made an algebra mistake.
 
Oops, thanks, Suppose to be 45/30 = 1.5.
So Car A will travel another 120 km before Car B catches up?

B)asks the distance Car B will be ahead of for 30 seconds after it passes Car A.
Car B 110 km/h(110 km per 3600 seconds), Car A 80 km/hr(80 km per 3600 seconds)
each second for Car B covers .031km and each second for Car A covers .022km
.031km * 30sec = .93 km.

So after 30 seconds, Car B will be .93 km ahead of Car A

Is this approach correct?
 
  • #10
ff4930 said:
Oops, thanks, Suppose to be 45/30 = 1.5.
So Car A will travel another 120 km before Car B catches up?

B)asks the distance Car B will be ahead of for 30 seconds after it passes Car A.
Car B 110 km/h(110 km per 3600 seconds), Car A 80 km/hr(80 km per 3600 seconds)
each second for Car B covers .031km and each second for Car A covers .022km
.031km * 30sec = .93 km.

So after 30 seconds, Car B will be .93 km ahead of Car A

Is this approach correct?
Not quite - .031 km/sec * 30 sec gives the distance B has traveled, but A isn't just standing still - it's covered some distance, too. The difference between the two tells you how far B is ahead of A.
 
  • #11
Cool! Thanks belliott, much obliged.
 

Similar threads

Problem with the power of car
  • · Replies 3 ·
Replies
3
Views
1K
How Do You Calculate Race Car Acceleration and Power Output?
  • · Replies 20 ·
Replies
20
Views
2K
Help with question on motion: Avoiding a rear-end collision
  • · Replies 6 ·
Replies
6
Views
3K
Distance travelled by a car considering only air friction?
  • · Replies 4 ·
Replies
4
Views
2K
Determining the interval between two F1 cars
  • · Replies 30 ·
2
Replies
30
Views
3K
Kinematics: Motion in One Direction: Car Chase
  • · Replies 9 ·
Replies
9
Views
3K
Calculating Time and Acceleration in Constantly Accelerating Cars
  • · Replies 30 ·
2
Replies
30
Views
10K
How Long Does It Take for One Car to Overtake Another?
  • · Replies 5 ·
Replies
5
Views
2K
Car Rounding Curve: Acceleration Calculation with C=2piR
  • · Replies 5 ·
Replies
5
Views
2K
Motion with constant acceleration
  • · Replies 4 ·
Replies
4
Views
2K